heureka

Mrs Kumar paid $14 438 for 5 similar laptops and 4 similar DVD players.

If 1 laptop and 1 DVD player cost $2960 altogether,

what was the cost of each DVD player?

Let x = laptop
Let y = DVD player

\(\begin{array}{|lrcll|} \hline (1) & x+y &=& $2960 \quad & | \quad -y \\ & x &=& 2960 - y \\\\ (2) & 5x+4y &=& $14438 \quad & | \quad x= 2960 - y \\ & 5\cdot (2960 - y) + 4y &=& 14438 \\ & 5\cdot 2960 - 5y + 4y &=& 14438 \\ & 14800 - 5y + 4y &=& 14438 \\ & 14800 - y &=& 14438 \quad & | \quad \cdot (-1) \\ & -14800 + y &=& -14438 \quad & | \quad +14800 \\ & y &=& 14800 - 14438 \\ & \mathbf{y} & \mathbf{=} & \mathbf{$362 } \\\\ & x &=& 2960 - y \quad & | \quad y= 362 \\ & x &=& 2960 - 362 \\ & \mathbf{x} & \mathbf{=} & \mathbf{$2598 } \\ \hline \end{array}\)

Each DVD player cost $362

The sum of three numbers is 7656.

The first number is 1099 more than the second number.

The third number is 877 less than the second number.

Find the value of the second number.

Let n₁ = Number 1

Let n₂ = Number 2

Let n₃ = Number 3

\(\small{ \begin{array}{|rcll|} \hline n_1+n_2+n_3 &=& 7656 \quad n_1 = 1099 +n_2 \quad n_3 = n_2-877 \\ (1099 +n_2)+n_2+(n_2-877) &=& 7656 \\ 1099 +n_2+n_2 +n_2-877 &=& 7656 \\ 1099 +3n_2-877 &=& 7656 \\ 3n_2+ 1099 -877 &=& 7656 \\ 3n_2+ 222 &=& 7656 \quad & | \quad -222 \\ 3n_2 &=& 7656 - 222 \\ 3n_2 &=& 7434 \quad & | \quad :3 \\ n_2 &=& \frac{7434}{3} \\ \mathbf{n_2} & \mathbf{=} & \mathbf{2478} \\\\ n_1 &=& 1099 +n_2 \quad & | \quad n_2 = 2478 \\ n_1 &=& 1099 +2478 \\ \mathbf{n_1} & \mathbf{=} & \mathbf{3577} \\\\ n_3 &=& -877 +n_2 \quad & | \quad n_2 = 2478 \\ n_3 &=& -877 +2478 \\ \mathbf{n_3} & \mathbf{=} & \mathbf{1601} \\ \hline \end{array} }\)

The second number is 2478

Integral von e^(-x^2)

Die Exponentialfunktion als Reihe: \(e^{x}=\sum \limits_{k=0}^\infty{\frac {x^{k}}{k!}}={\frac {x^{0}}{0!}}+{\frac {x^{1}}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots \)

Die Exponentialfunktion als Reihe von: \(\begin{array}{|rcll|} \hline e^{(-x^2)} &=& \sum \limits_{k=0}^\infty {\frac {(-x^2)^{k}} {k!} } \\ &=& \sum \limits_{k=0}^\infty {\frac {(-1)^kx^{2k} }{k!}} \\ \hline \end{array}\)

Jeder Summand wird für sich integriet:

\(\begin{array}{|rcll|} \hline \int e^{(-x^2)}\ dx &=&\int \Big( \sum \limits_{k=0}^\infty {\frac {(-1)^kx^{2k} }{k!}} \Big) \ dx \\ &=& \sum \limits_{k=0}^\infty\frac{(-1)^k x^{1+2k}}{(1+2k)k!} \\ \hline \end{array}\)

Consider the relation y = −5|x − 4| + 7.

What are the coordinates of the vertex, V, and which direction does it open?

The line represented by y = 2x + 1 and a line perpendicular to it intersect at R(2, 5).

What is the equation of the perpendicular line?

\(\begin{array}{|rcll|} \hline y &=& 2x + 1 \quad m = 2 \quad & \quad m_{\perp} = -\frac{1}{m} = - \frac12 \\\\ y_{\perp} &=& -\frac12\cdot x + b \quad & \quad R=(2,5) \\ 5 &=& -\frac12\cdot 2 + b \\ 5 &=& -1 + b \\ 6 &=& b \\ y_{\perp} &=& -\frac12\cdot x + 6 \\ \hline \end{array} \)

The equation of the perpendicular line is \(y = -\frac12\cdot x + 6\)

A stock market analyst observes the following for the price of two stocks that he owns, one of which is growing in a linear fashion (arithmetic) and the other of which is growing at an exponential rate (geometric).

Stock A:

\(a_n = 15n+50 \\ a_{40} = 15\cdot 40 + 50 \\ a_{40} = $650 \)

Stock B:

\(a_n = 13\cdot 1-11^{n-1} \\ a_{40} = 13\cdot 1-11^{39} \\ a_{40} = $761.27 \)

Assuming these stock values continue to grow in the same manner until retirement,

which stock option will be worth more in 40 years,

and how much more (rounded to the nearest cent) will this stock be worth per share?

Stock B;  $761.27- $650 = $ 111.27

What is the general-term equation, an, for the arithmetic sequence below,

and what is the 23rd term of this sequence?
\(\begin{array}{|l|r|r|r|} \hline n & 1 & 2 & 3 \\ \hline a_n & 2 & -3& -8 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)\cdot d \quad & | \quad a_1 = 1 \\ && \quad & | \quad d = a_2-a_1\\ && \quad & | \quad = -3-2 \\ && \quad & | \quad = -5 \\ a_n &=& 2 + (n-1)(-5) \\ a_n &=& 2 -5n + 5 \\ a_n &=& 7 -5n \\\\ a_{23} &=& 7 - 5\cdot 23 \\ a_{23} &=& -108 \\ \hline \end{array}\)

One endpoint of a line segment is at (−3, 1).

The line is bisected by placing the midpoint of the line segment at (5, −3).

What are the coordinates of the other endpoint?

\(\begin{array}{|lrcll|} \hline \frac{-3+x}{2} &=& 5 \\ -3+x &=& 10 \\ x &=& 13 \\\\ \frac{1+y}{2} &=& -3 \\ 1+y &=& -6 \\ y &=& -7 \\ \hline \end{array}\)

The coordinates of the other endpoint is (13, -7)

How many terms are in the arithmetic sequence 5, 1, −3, …, −99?
Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference

\(\begin{array}{|rcll|} \hline 5, 1, -3, \ldots, -99 \\ a_1 &=& 5 \\ a_2 &=& a_1 + d \\ d &=& a_2 - a_1 \\ d &=& 1-5 \\ d &=& -4 \\\\ a_n &=& a_1 + (n-1)\cdot d \quad & | \quad a_1 = 5 \quad a_n=-99 \quad d = -4 \\ -99 &=& 5 + (n-1)\cdot (-4) \quad & | \quad -5 \\ -104 &=& (n-1)\cdot (-4) \quad & | \quad : (-4) \\ \frac{-104}{-4} &=& n-1 \\ \frac{104}{4} &=& n-1 \\ 26 &=& n-1 \quad & | \quad +1 \\ 27 &=& n \\ \hline \end{array}\)

In the arithmetic sequence 5, 1, -3, …, -99 are 27 terms.

Hallo Heureka, muss nicht der Term \(-\sqrt{\frac{2}{3}}\) durchgängig beibehalten werden? Wegen \(f\Big(-\sqrt{\frac{2}{3}}\Big)\)?

Hallo asinus,

ich denke nein, denn:

\(\begin{array}{|rcll|} \hline y=f(x) &=& x^3-2x \\ y &=& x\cdot (x^2-2) \quad & | \quad x = -\sqrt{\frac23} \\ y &=& \Big(-\sqrt{\frac23} \Big) \cdot \Big[ \left(-\sqrt{\frac23}\right)^2-2 \Big] \\ y &=& \Big(-\sqrt{\frac23} \Big) \cdot \Big( \frac23-2 \Big) \\ y &=& \Big(-\sqrt{\frac23} \Big) \cdot \Big( -\frac43 \Big) \\ y &=& \Big(\sqrt{\frac23} \Big) \cdot \Big( \frac43 \Big) \\ y &=& \frac{4}{3}\cdot \sqrt{\frac23} \\ \hline \end{array} \)

Gruß Heureka!